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Other Aspects of Ionic Equilibria 1
Solubility of Salts - Ksp
We now focus on another aqueous equilibrium system, slightly
soluble salts.
These salts have a Solubility Product Constant, Ksp.(We saw this
in 1B with the sodium tetraborate solubility lab.)
For example, Ksp is defined for Cu(OH)2(s) as follows:
Cu(OH)2(s) Cu2+(aq) + 2 OH(aq)
Ksp=[Cu2+][OH]2 = 4.8x1020 at 25 C
For a list of Ksp values at 25C, refer to Appendix D-Table
D.3.
These slightly soluble salts dissolve until Ksp is satisfied. At
this point, we say the solution is saturated, and no more salt will
dissolve. In terms of Qsp we have the following possibilities:
1. Qsp < Ksp
2. Qsp = Ksp
3. Qsp > Ksp
Other Aspects of Ionic Equilibria 2
Ksp SolubilityWrite the chemical equilibrium and the Ksp
equilibrium-constant expression for the solubility of
Ca3(PO4)2.
Solubility and Ksp The solubility of salts (saturated solution)
are often expressed in one or more of the
following units with the temperature also specified:mol/L (molar
solubility)
g/L , g/mL, g/100 mL or mg/L (Note that these are NOT
densities.)ppm or ppb (parts per million or parts per billion) for
very insoluble salts.
The above units for solubility are directly related to Ksp
through equilibrium stoichiometry. Lets do some examples.
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Other Aspects of Ionic Equilibria 3
Solubility and Ksp1. What is the molar solubility of Ca3(PO4)2
at 25C given a Ksp of 2.0x10
29?
1.1. What is the concentration of calcium ions?
1.2. What is this solubility in g/L solution?
1.3. What is this solubility in parts per million (ppm) and
parts per billion (ppb)? (Assume a soln density of 1.0 g/mL)
ppm = mass of component in solntotal mass of soln x106
ppb = mass of component in solntotal mass of soln x109
Other Aspects of Ionic Equilibria 4
Ksp from Experimental DataA saturated solution of magnesium
hydroxide in water has a pH of 10.38. Calculate the Ksp for
magnesium hydroxide.
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Other Aspects of Ionic Equilibria 5
Relative Molar Solubility of Salts
The relative MOLAR solubility of salts (saturated solution) can
be determined by comparing Ksp values. The greater the Ksp the more
ions are in solution, hence the greater the molar solubility.
However, you can only directly compare salts that give
equivalent numbers of ions in solution.
For example, you can compare the Ksps of all salts with a 1:1
ion ratio like: AgBr, BaSO4 etc.
Or, you could compare Ksps of all salts with 1:2 and 2:1 ion
ratios like: BaCl2, Ag2SO4, etc.
But you cant compare the Ksp AgBr directly to the Ksp Ag2SO4 to
determine which is more soluble.
Which salt has a greater molar solubility soluble: AgBr or
BaSO4?
Other Aspects of Ionic Equilibria 6
Ksp Values and Solubility Calcs. LIMITATIONS EXIST!
Unfortunately, solubilities calculated using Ksp values
sometimes deviate appreciably from the experimentally measured
solubilities! Why?
Reason 1 of 2: Some salts do not completely dissociate 100% into
their respective ions. For example, PbCl2 exists in three different
forms in water, each with its own individual K value.
1.PbCl2(s) PbCl2(aq) (no ions) K = 1.1x10-3; S = 0.0011
2.PbCl2(s) PbCl +(aq) + Cl(aq) (two ions) K = 6.9x10-4; S =
0.0263.PbCl2(s) Pb2+(aq) + 2 Cl(aq) (all ions) K = 1.7x10-5; S =
0.016
The experimental solubility of PbCl2 is 0.036 M. Over twice the
value predicted by Ksp alone.
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Other Aspects of Ionic Equilibria 7
Reason 2: Some anions in salts are strongly basic. They react
with water in a Kb equilibrium. This reduces the anion
concentration available to satisfy Ksp. This in turn increases the
solubility of the salt.
The three common ions that are basic enough to have considerable
reaction with water are: S2-, CO3
2-, and PO43-.
For example consider Ca3(PO4)2:The Ksp equilibrium is: Ca3(PO4)2
3 Ca+(aq) + 2 PO43(aq) Ksp = 2.0x1029
Write the chemical reaction for the Kb equilibrium (hydrolysis)
for PO43 (aq):
Kb = ? (How do you determine the value?)
The experimental solubility of Ca3(PO4)2 is somewhat higher than
predicted by Ksp alone because some phosphate ion is removed from
solution through the Kb reaction.Ksp values are still useful for
estimating solubilities, predicting trends and predicting relative
solubilities. We just need to keep in mind that they have
limitations!
We previously determined the molar solubility of Ca3(PO4)2(s)
using Ksp alone. Now lets determine its solubility including Kb
with Ksp to find Knet.
Ksp Values and Solubility Calcs. LIMITATIONS
Other Aspects of Ionic Equilibria 8
The Ksp of Sulfide Salts INCLUDES Kb
Since the sulfide ion is very basic, Kb significantly increases
the solubility of sulfide salts. For the metal sulfides, the Ksp
values are actually Knet for the overall equilibrium that includes
two processes:
1) the dissolution of the metal sulfide
2) the Kb hydrolysis reaction of the the sulfide ion.
(See the footnote at the bottom of Table D.3 in our text
book)
For CuS write the chemical reaction for each of the above
equilibria, and for the net equilibrium:
1. K1
2. Kb >> 1
Net: Ksp = K1*Kb = 6x10-37 (Table D.3)
Determine the pH of a saturated CuS solution.
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Other Aspects of Ionic Equilibria 9
Solubility of Salts and the Common Ion Effect
Like the percent ionization of weak acids and bases, the
solubility of salts is influenced by the presence of a common ion.
These salts are less soluble when a common ion is present just like
a weak acids ionization is limited in the presence of significant
conjugate base.
Play Movie
1. The Ksp of silver iodide in water is 8.3x1017 M. Calculate
the molar solubility of silver iodide in:1. Pure water2. 0.010 M
NaI (Common Ion Effect)
Other Aspects of Ionic Equilibria 10
Solubility of Salts and the Common Ion Effect
1. The barium ion, Ba2+(aq), is poisonous when ingested. The
lethal does in mice is about 12 mg Ba2+ per kg of body mass.
Despite this fact, BaSO4 is widely used in medicine to obtain X-ray
images of the gastrointestinal tract since Ba is a very good x-ray
absorbing element.
a) Explain why BaSO4(s) is safe to ingest, even though Ba2+(aq)
is poisonous.
b) Calculate the concentration of Ba2+, in milligrams per liter,
in saturated BaSO4(aq) at 25C.
c) Soluble MgSO4 is often mixed with BaSO4 when ingested. What
function does the MgSO4 serve?
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Other Aspects of Ionic Equilibria 11
Solubility of Salts with pH AdjustmentsWe have already seen that
basic salts containing S2-, CO3
2-, and PO43- ions are more soluble in water
than expected because of the reaction of the basic anion with
water through Kb.
All salts that contain a basic anion will have their solubility
increased in an acidic solution. In acidic solutions the basic
anion reacts with the acid, forcing more of the salt to dissolve to
reach equilibrium.
For example, PbF2 and Mg(OH)2+ are practically insoluble in
water. However, they are very soluble in dilute acids.
Play Movie
Other Aspects of Ionic Equilibria 12
Solubility of Salts with pH Adjustments
1. Which of the following slightly soluble salts will be more
soluble in acidic solution than pure water? For those where the
solubility increases, write the net ionic chemical reaction that
occurs when a strong acid is present in solution:
a) Al(OH)3(s)
b) BaSO4(s)
c) BaC2O4(s)
d) PbCl2
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Other Aspects of Ionic Equilibria 13
Solubility and pH1. Calculate the molar solubility of iron (II)
hydroxide (Ksp = 7.9x10-16) in
a) A solution buffered at pH=7.00:
b) What could you say about the solubility of iron (II)
hydroxide in a solution buffered at pH=10.00
Other Aspects of Ionic Equilibria 14
Solubility and pH1. A chief component in marble is calcium
carbonate. Marble has been widely used for
statues and ornamental work on buildings, but marble is readily
attacked by acids. Assume that the overall reaction that occurs in
a dilute acid is
CaCO3(s) + H3O+(aq) Ca2+(aq) + HCO3(aq) + H2O(l)Determine the
equilibrium constant for this reaction* and then determine the
solubility of calcium carbonate in a buffer with pH = 5.6 (pH stays
constant).
*Hint: What combination of reactions yields the overall
reaction?
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Other Aspects of Ionic Equilibria 15
Solubility of Salts and pH-ApplicationsA principle cause of
sinkhole formation: dissolution of limestone (CaCO3) by
naturally acidic rain water as it percolates through the
bedrock. Voids in the bed rock can cause sudden collapse of the
overlying ground.
Dissolved CaCO3
Sinkhole forming
Sinkhole collapses
Other Aspects of Ionic Equilibria 16
Sinkhole Formation
The large sinkhole shown here has destroyed several buildings
and part of a highway.
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Other Aspects of Ionic Equilibria 17
1. Two of the main crystalline components of kidney stones are
calcium phosphate and calcium oxalate. What happens to the
solubility of these compounds as pH is increased?
2. Tooth enamel consists mainly of a mineral called
hydroxyapatite, Ca5(PO4)3(OH)(s), Ksp=6.8x1037, that is insoluble
in pure water. When acids dissolve tooth enamel, the result is
tooth decay. Write the net ionic equation that occurs between
hydroxyapatite and H3O+(aq). Note, the phosphate ions are also
pronated along with the hydroxide.
3. Fluoridation of water and the use of fluoride toothpaste
causes the OH ion in hydroxyapatite to be replaced with F forming
Ca5(PO4)3F (Ksp=6.8x1060). Suggest a reason why fluoride helps
prevent tooth decay.
Solubility of Salts and pH-Applications
Other Aspects of Ionic Equilibria 18
Complex Ion Formation, KfComplex ions are ions that usually
contain a transition metal cation and one or more ligands. A ligand
is either a neutral molecule or an anion that bonds to the metal
cation through a Lewis acid-base reaction:
Metal Ion: Lewis acid - LIMITING REACTANT
Ligand: Lewis Base - EXCESS REACTANT
Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) Kf = 1.7x107 = ? (Write the
expression)
What can you conclude about the favorability of complex ion
formation?
What can you conclude about the free (uncomplexed) metal cation
concentration in solutions where a complex ion can form and the
ligand is in excess?
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Other Aspects of Ionic Equilibria
Determine the concentration of free (uncomplexed) silver ions in
solution when 5.0 mL of a 6.0M ammonia solution (ligand) is added
to 45.0 mL of a 0.10 M silver nitrate solution (metal cation).
19
Complex Ion Formation, Kf
Other Aspects of Ionic Equilibria 20
Complex Ion Formation and SolubilityFormation of complex ions is
one way to dissolve insoluble salts. Consider the following
sequence of additions to a solution containing Ag+ ions.
Precipitation of AgCl
Complex ion formation
Precipitation of AgBr
Complex ion formation
AgCl is precipitated by adding NaCl(aq) to AgNO3(aq)
The AgCl is dissolved by adding excess aqueous NH3
The silver-ammonia complex ion is changed to insoluble AgBr on
addition of NaBr(aq).
The AgBr is dissolved by adding excess Na2S2O3(aq).
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Other Aspects of Ionic Equilibria 21
Complex Ion Formation and Solubility
Using equilibrium constants (Ksp and Kf) we can show why each
process in the previous illustration can be made favorable.
Step (a): Precipitation of AgCl (Ksp = 1.8x10-10)
Ag+(aq) + Cl(aq) AgCl(s) Kppt = ?
What concentration of Cl(aq) is required to precipitate 99.9% of
the Ag+(aq) ions from a 0.10 M AgNO3(aq) solution. What mass of
NaCl (molar mass = 58.44 g/mol) must be added to 100.0 mL of the
solution to accomplish this? (Assume that the volume change of the
solution is negligible upon addition of the NaCl.)
Other Aspects of Ionic Equilibria 22
Complex Ion Formation and SolubilityStep (b): AgCl(s) dissolves
when NH3(aq) is added:
AgCl(s) Ag+(aq) + Cl-(aq) Ksp = 1.8x1010
Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) Kf = 1.7x107
AgCl(s) + 2 NH3(aq) Ag(NH3)2+(aq) + Cl(aq) Knet = ?A high
concentration of the ligand, NH3(aq), in solution ensures that
almost all the AgCl(s) will dissolve. This is an application of Le
Chateliers Principle.
This is a key point for dissolving insoluble salts by formation
of a complex ion:Salt + Ligand > Complex ion + Anion Knet =
Ksp*Kf
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Other Aspects of Ionic Equilibria 23
Step (b): What is the molar solubility of AgCl(s) ina) pure
H2O(l):
b) a solution where the concentration of NH3(aq) begins at 0.10
M:
Calculations: Complex Ion Formation and Solubility
Other Aspects of Ionic Equilibria 24
Step (c): Ag(NH3)2+(aq) precipitates as AgBr(s) when NaBr(aq) is
added.
Find Knet for this system.
Determine the concentration of complexed ion still in solution
where the Ag(NH3)2+(aq) and NaBr(aq)
concentrations both begin at 0.10M.
Calculations: Complex Ion Formation and Solubility
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Other Aspects of Ionic Equilibria 25
Step (d): AgBr(s) dissolves as Na2S2O3(aq) is added. (Write the
net ionic equation for the reaction below.)
Crystals of AgBr can be removed from black and white
photographic film by reacting the AgBr(s) with aqueous sodium
thiosulfate. In order to dissolve 2.5 g of AgBr in 1.00 L of
solution, what concentration of thiosulfate ion is needed in
solution at equilibrium?
Now calculate how many grams of solid Na2S2O3 (molar mass =
158.11g/mol ) must be added to dissolve the AgBr.
Complex Ion Formation and Solubility
Other Aspects of Ionic Equilibria 26
Amphoterism of Metal Oxides and HydroxidesDefinition of
amphoteric: An amphoteric substance is slightly soluble in water,
but soluble in either acidic or basic solutions.Example Al3+:
(waters of hydration omitted for clarity)1. Al(OH)3(s) + 3 H+(aq)
Al3+(aq) + 3 H2O(l)2. Al3+(aq) + OH(aq) AlOH2+(aq)3. AlOH2+(aq) +
OH(aq) Al(OH)2+(aq)4. Al(OH)2+(aq) + OH(aq) Al(OH)3(s)5. Al(OH)3(s)
+ OH(aq) Al(OH)4(aq)
} Weakly basic solution, 7 < pH < 9 Strongly basic
solution, pH > 10 Basic solution, pH 10
Other amphoteric oxides and hydroxides:
Cr3+, Zn2+, and Sn2+Each will complex a different # of hydroxide
ions - you will
have to check a table of Kf values.
pH = 10pH 10
Acidic solution, pH < 7
ppt
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Other Aspects of Ionic Equilibria 27
Zinc hydroxide is insoluble in water (Ksp= 3.0x1016), but
amphoteric. Write net ionic equations to show
why Zn(OH)2 is readily soluble in the following solutions:
(a) CH3COOH(aq)
(b) NH3(aq): Zn2+ ions form the following complex ion:
[Zn(NH3)4]
2+ with a Kf = 4.1x108
(c) OH(aq) Zn2+ ions form the following complex ion: [Zn(OH)4]
2- with a Kf = 4.6x1017
(c.1) What is the minimum pH required to dissolve 1.5 g of
Zn(OH)2 in a liter of solution?
Amphoterism of Metal Oxides and Hydroxides
Other Aspects of Ionic Equilibria 28
Precipitation Reactions and Separation of Ions
Recall from previous slides about solubility: Qsp < Ksp :
unsaturated soln. - no precipitation.
Qsp = Ksp: saturated soln. - equilibrium.
Qsp > Ksp: supersaturated soln. - a ppt. should form.
Using Ksp values:
1. We can determine if a ppt will form when two solutions are
mixed that contain a cation and an anion that can combine to form a
slightly soluble salt.
2. We can determine if different cations in solution can be
quantitatively separated from each other by selective
precipitation. In other words, we can determine if one cation can
be removed from the solution as a precipitated salt before a second
cation in solution is also precipitated.
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Other Aspects of Ionic Equilibria 29
Single Precipitation ReactionsCalcium ion triggers clotting of
blood. Therefore, when blood is donated the receiving bag contains
sodium oxalate to precipitate the Ca2+, and thus prevent clotting.
Typically, blood contains 9.7x10-5 g Ca2+/mL. To remove the calcium
ions, a medical technologist treats a 104-mL blood sample with
100.0 mL of 0.155 M Na2C2O4. Calculate the [Ca2+] left in solution
after the treatment. (Ksp for calcium oxalate = 2.6x10-9)
Other Aspects of Ionic Equilibria 30
Single Precipitation Reactions
1. Show that a precipitate of Mg(OH)2 will form in an aqueous
solution that is 0.350 M MgCl2 and 0.750 M NaOH.
2. Show that a precipitate of Mg(OH)2 will also form in an
aqueous solution that is 0.350 M MgCl2 and 0.750 M NH3. Hint: Find
the [OH
-] produced by the hydrolysis of NH3. Is it enough to ppt the
Mg2+ ion?
2.1. Explain why the Mg(OH)2 precipitate can be prevented from
forming in (2) by adding NH4Cl to the solution containing
ammonia.
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Other Aspects of Ionic Equilibria 31
Single Precipitation ReactionsCalculate the minimum pH needed to
precipitate Mn(OH)2 so completely that the concentration of Mn2+ is
less than 1 g per liter, that is 1 part per billion (ppb).
Other Aspects of Ionic Equilibria 32
Selective Precipitation of CationsDifferences in molar
solubilities between compounds containing a common ion can be used
to selectively precipitate one ion from solution leaving the other
ion in solution. We can calculate the amount of the common ion
needed to reach saturation for the most soluble ionic salt.
Addition of just a little less of the common ion will insure the
most complete separation possible.
Quantitative, or complete, separation is considered possible if
99.9% of the least soluble salt precipitates before any of the most
soluble salt starts to precipitate. Typically, the added reagent
(containing the common ion) is quite concentrated so that its
addition does not appreciable change the volume of the solution and
dilute the solution containing the cations to be separated.
Solution of cationsAdd ppt agent Least soluble
cation ppts first
Centrifuge and decant Selective ppt complete
A 400.0 mL solution has [Ba2+] = 0.0040 M and [Sr2+] = 0.0010
M.A 0.800 M Na2SO4(aq) solution is added drop-wise.
Which ion, Ba2+ or Sr2+, will precipitate first?Ksp for BaSO4 is
1.1x10-10, Ksp for SrSO4 is 3.4x10-7
Can the two cations be quantitatively (99.9%) separated by
selective precipitation?
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Other Aspects of Ionic Equilibria
ApplicationsPrecipitation Reactions:
33
Precipitation of Mg2+ as as Mg(OH)2 from sea water is the source
of Mg metal.
Titration of solutions containing Cl(aq) with AgNO3 to
quantitatively determine [Cl]. For this to work,we must be sure
that almost all of the Cl precipitates as we add the AgNO3. How can
we be sure of this?
Determination of the amount of SO42 in solution by precipitating
as BaSO4(s). The BaSO4 formed is filtered off, dried and weighed.
For this to work, the precipitation of SO42 must be complete
(99.9%). How can we be sure of this?
Complex Ion Formation: Complex ion formation can be used to
extract gold from low-grade gold containing rock. The
formation constant of Au(CN)2 is very large. A very small
concentration of Au+ ions are formed through oxidation of Au. This
oxidation takes place in the presence of CN ions. Complex ion
formation removes the Au+ ions from solution, so more are formed.
As a result, even though Au is not oxidized normally by air,
bubbling air through a suspension of Au containing ore in the
presence of CN leads to formation of a solution of the complex
ion:
4Au(s) + 8CN(aq) + O2(g) +2H2O(l) 4Au(CN)2(aq) + 4OH(aq)The
resulting solution is filtered and the Au+ reduced to Au(s).
Aluminum ore contains Fe2O3 impurities along with the desired
aluminum hydroxide, Al2O3. A strong base is added to dissolve the
Al2O3 as the Al(OH)4 complex ion (Kf = 3x1033). The iron (III) ion
does not form a complex ion with hydroxide, therefore the Fe2O3
does not dissolve. The resulting solution is filtered and acid is
added to the filtrate to precipitate Al(OH)3. The filtrate must not
be made too acidic or the Al(OH)3 will redissolve!